### 3.951 $$\int e^{c+d x} \coth (a+b x) \text{csch}^2(a+b x) \, dx$$

Optimal. Leaf size=113 $\frac{4 e^{2 a+x (2 b+d)+c} \, _2F_1\left (2,\frac{1}{2} \left (\frac{d}{b}+2\right );\frac{1}{2} \left (\frac{d}{b}+4\right );e^{2 (a+b x)}\right )}{2 b+d}-\frac{8 e^{2 a+x (2 b+d)+c} \, _2F_1\left (3,\frac{1}{2} \left (\frac{d}{b}+2\right );\frac{1}{2} \left (\frac{d}{b}+4\right );e^{2 (a+b x)}\right )}{2 b+d}$

[Out]

(4*E^(2*a + c + (2*b + d)*x)*Hypergeometric2F1[2, (2 + d/b)/2, (4 + d/b)/2, E^(2*(a + b*x))])/(2*b + d) - (8*E
^(2*a + c + (2*b + d)*x)*Hypergeometric2F1[3, (2 + d/b)/2, (4 + d/b)/2, E^(2*(a + b*x))])/(2*b + d)

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Rubi [A]  time = 0.272688, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.091, Rules used = {5511, 2251} $\frac{4 e^{2 a+x (2 b+d)+c} \, _2F_1\left (2,\frac{1}{2} \left (\frac{d}{b}+2\right );\frac{1}{2} \left (\frac{d}{b}+4\right );e^{2 (a+b x)}\right )}{2 b+d}-\frac{8 e^{2 a+x (2 b+d)+c} \, _2F_1\left (3,\frac{1}{2} \left (\frac{d}{b}+2\right );\frac{1}{2} \left (\frac{d}{b}+4\right );e^{2 (a+b x)}\right )}{2 b+d}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(c + d*x)*Coth[a + b*x]*Csch[a + b*x]^2,x]

[Out]

(4*E^(2*a + c + (2*b + d)*x)*Hypergeometric2F1[2, (2 + d/b)/2, (4 + d/b)/2, E^(2*(a + b*x))])/(2*b + d) - (8*E
^(2*a + c + (2*b + d)*x)*Hypergeometric2F1[3, (2 + d/b)/2, (4 + d/b)/2, E^(2*(a + b*x))])/(2*b + d)

Rule 5511

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(G_)[(d_.) + (e_.)*(x_)]^(m_.)*(H_)[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol]
:> Int[ExpandTrigToExp[F^(c*(a + b*x)), G[d + e*x]^m*H[d + e*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
IGtQ[m, 0] && IGtQ[n, 0] && HyperbolicQ[G] && HyperbolicQ[H]

Rule 2251

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[(a^p*G^(h*(f + g*x))*Hypergeometric2F1[-p, (g*h*Log[G])/(d*e*Log[F]), (g*h*Log[G])/(d*e*Log[F]) + 1, Simplify
[-((b*F^(e*(c + d*x)))/a)]])/(g*h*Log[G]), x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] ||
GtQ[a, 0])

Rubi steps

\begin{align*} \int e^{c+d x} \coth (a+b x) \text{csch}^2(a+b x) \, dx &=\int \left (\frac{8 e^{2 a+c+(2 b+d) x}}{\left (-1+e^{2 (a+b x)}\right )^3}+\frac{4 e^{2 a+c+(2 b+d) x}}{\left (-1+e^{2 (a+b x)}\right )^2}\right ) \, dx\\ &=4 \int \frac{e^{2 a+c+(2 b+d) x}}{\left (-1+e^{2 (a+b x)}\right )^2} \, dx+8 \int \frac{e^{2 a+c+(2 b+d) x}}{\left (-1+e^{2 (a+b x)}\right )^3} \, dx\\ &=\frac{4 e^{2 a+c+(2 b+d) x} \, _2F_1\left (2,\frac{1}{2} \left (2+\frac{d}{b}\right );\frac{1}{2} \left (4+\frac{d}{b}\right );e^{2 (a+b x)}\right )}{2 b+d}-\frac{8 e^{2 a+c+(2 b+d) x} \, _2F_1\left (3,\frac{1}{2} \left (2+\frac{d}{b}\right );\frac{1}{2} \left (4+\frac{d}{b}\right );e^{2 (a+b x)}\right )}{2 b+d}\\ \end{align*}

Mathematica [A]  time = 1.54049, size = 159, normalized size = 1.41 $-\frac{e^{c-\frac{a d}{b}} \left (d^2 e^{\left (\frac{d}{b}+2\right ) (a+b x)} \, _2F_1\left (1,\frac{d}{2 b}+1;\frac{d}{2 b}+2;e^{2 (a+b x)}\right )+d (2 b+d) e^{d \left (\frac{a}{b}+x\right )} \, _2F_1\left (1,\frac{d}{2 b};\frac{d}{2 b}+1;e^{2 (a+b x)}\right )+(2 b+d) e^{d \left (\frac{a}{b}+x\right )} \left (d \coth (a+b x)+b \text{csch}^2(a+b x)\right )\right )}{2 b^2 (2 b+d)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(c + d*x)*Coth[a + b*x]*Csch[a + b*x]^2,x]

[Out]

-(E^(c - (a*d)/b)*((2*b + d)*E^(d*(a/b + x))*(d*Coth[a + b*x] + b*Csch[a + b*x]^2) + d*(2*b + d)*E^(d*(a/b + x
))*Hypergeometric2F1[1, d/(2*b), 1 + d/(2*b), E^(2*(a + b*x))] + d^2*E^((2 + d/b)*(a + b*x))*Hypergeometric2F1
[1, 1 + d/(2*b), 2 + d/(2*b), E^(2*(a + b*x))]))/(2*b^2*(2*b + d))

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Maple [F]  time = 0.1, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{dx+c}}\cosh \left ( bx+a \right ) \left ({\rm csch} \left (bx+a\right ) \right ) ^{3}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(exp(d*x+c)*cosh(b*x+a)*csch(b*x+a)^3,x)

[Out]

int(exp(d*x+c)*cosh(b*x+a)*csch(b*x+a)^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -48 \, b d^{2} \int \frac{e^{\left (d x + c\right )}}{48 \, b^{3} - 44 \, b^{2} d + 12 \, b d^{2} - d^{3} +{\left (48 \, b^{3} - 44 \, b^{2} d + 12 \, b d^{2} - d^{3}\right )} e^{\left (8 \, b x + 8 \, a\right )} - 4 \,{\left (48 \, b^{3} - 44 \, b^{2} d + 12 \, b d^{2} - d^{3}\right )} e^{\left (6 \, b x + 6 \, a\right )} + 6 \,{\left (48 \, b^{3} - 44 \, b^{2} d + 12 \, b d^{2} - d^{3}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 4 \,{\left (48 \, b^{3} - 44 \, b^{2} d + 12 \, b d^{2} - d^{3}\right )} e^{\left (2 \, b x + 2 \, a\right )}}\,{d x} + \frac{4 \,{\left (12 \, b d e^{c} +{\left (24 \, b^{2} e^{c} - 10 \, b d e^{c} + d^{2} e^{c}\right )} e^{\left (4 \, b x + 4 \, a\right )} -{\left (24 \, b^{2} e^{c} + 2 \, b d e^{c} - d^{2} e^{c}\right )} e^{\left (2 \, b x + 2 \, a\right )}\right )} e^{\left (d x\right )}}{48 \, b^{3} - 44 \, b^{2} d + 12 \, b d^{2} - d^{3} -{\left (48 \, b^{3} - 44 \, b^{2} d + 12 \, b d^{2} - d^{3}\right )} e^{\left (6 \, b x + 6 \, a\right )} + 3 \,{\left (48 \, b^{3} - 44 \, b^{2} d + 12 \, b d^{2} - d^{3}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 3 \,{\left (48 \, b^{3} - 44 \, b^{2} d + 12 \, b d^{2} - d^{3}\right )} e^{\left (2 \, b x + 2 \, a\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*cosh(b*x+a)*csch(b*x+a)^3,x, algorithm="maxima")

[Out]

-48*b*d^2*integrate(e^(d*x + c)/(48*b^3 - 44*b^2*d + 12*b*d^2 - d^3 + (48*b^3 - 44*b^2*d + 12*b*d^2 - d^3)*e^(
8*b*x + 8*a) - 4*(48*b^3 - 44*b^2*d + 12*b*d^2 - d^3)*e^(6*b*x + 6*a) + 6*(48*b^3 - 44*b^2*d + 12*b*d^2 - d^3)
*e^(4*b*x + 4*a) - 4*(48*b^3 - 44*b^2*d + 12*b*d^2 - d^3)*e^(2*b*x + 2*a)), x) + 4*(12*b*d*e^c + (24*b^2*e^c -
10*b*d*e^c + d^2*e^c)*e^(4*b*x + 4*a) - (24*b^2*e^c + 2*b*d*e^c - d^2*e^c)*e^(2*b*x + 2*a))*e^(d*x)/(48*b^3 -
44*b^2*d + 12*b*d^2 - d^3 - (48*b^3 - 44*b^2*d + 12*b*d^2 - d^3)*e^(6*b*x + 6*a) + 3*(48*b^3 - 44*b^2*d + 12*
b*d^2 - d^3)*e^(4*b*x + 4*a) - 3*(48*b^3 - 44*b^2*d + 12*b*d^2 - d^3)*e^(2*b*x + 2*a))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\cosh \left (b x + a\right ) \operatorname{csch}\left (b x + a\right )^{3} e^{\left (d x + c\right )}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*cosh(b*x+a)*csch(b*x+a)^3,x, algorithm="fricas")

[Out]

integral(cosh(b*x + a)*csch(b*x + a)^3*e^(d*x + c), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*cosh(b*x+a)*csch(b*x+a)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cosh \left (b x + a\right ) \operatorname{csch}\left (b x + a\right )^{3} e^{\left (d x + c\right )}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*cosh(b*x+a)*csch(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(cosh(b*x + a)*csch(b*x + a)^3*e^(d*x + c), x)